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Re: [Phys-L] finding Q from band width?



On 10/7/2013 10:12 PM, Bernard Cleyet wrote:
I'm attempting to find the Q of a clock pendulum while free from the escapement. (crutch and crown removed, so only the spring suspension in use) using the bandwidth method, i.e. Q = "natural" or resonance frequency / FWHPower, no?

I'm driving it E-magnetically using a PM waxed the the edge of the bob (lenticular) and a coil waxed to the case opposite the PM. Not only is this asymmetrical, but also the force is a strong (4th power?) function of the separation (near field). Therefore, I vary the drive to maintain the same amplitude, which is measured by the blocking time of a photogate at ~ BDC, as I manually change the drive frequency in 0.001Hz steps. I assume the drive power is the square of the applied EMF. The undriven frequency agrees w/ the minE drive frequency (~ 5%) [0.565Hz]


So if the min drive E is 0.3V^2 then the diff between the two frequencies at 0.6V^2 is the FWHP, yes??

The result is ~ 1k, which agrees ~ with the ring down method.

bc

I read BCs note given above carefully, several times.
I believe that he is asking a question, signaled by a sentence ending in "yes???"
I understand that this sentence could be represented like this:

If an energy increment, applied impulsively at each cycle to a pendulum moving at the frequency which needs least input energy (F1) is taken to be proportional to the square of the voltage value applied to a driving electromagnet, is it true to suppose that applying twice the value of this drive energy in order to maintain the previously observed amplitude at some higher and lower frequency (F3,F2) allows one to calculate the Quality Factor (Q) of the pendulum setup under consideration, by means of the relationship Q = F1/(F3 - F2)

In the telegraphic spirit in which the question was couched, I offer my own humble response:
no.

Sincerely


Brian Whatcott Altus OK